Question

A hydraulic machine exerts a force of $900N$ on a piston of diameter $1.80cm$. The output force is exerted on a piston of diameter $36cm$. What will be the output force?

Answer 1

Given,

$F_1=900N{F}_2=xN$

$r_1=0.9cm{r}_2=18cm$

From Pascals law, we have $P_1=P_2$

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

$\Rightarrow F_2=F_1\times \frac{r_2^2}{r_1^2}$

$F_2=\frac{90000\times 18\times 18}{0.9\times 0.9}$

$F_2=36\times {10}^{4}N$